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A084699
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Composite integers j such that binomial(2*j,j) == 2^j (mod j).
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0
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12, 30, 56, 424, 992, 16256, 58288, 119984, 356992, 1194649, 9973504, 12327121, 13141696, 22891184, 67100672, 233850649
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OFFSET
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1,1
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COMMENTS
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If p is prime, binomial(2*p,p) == 2^p (mod p).
a(17) > 10^9.
Theorem. Let j = (2^k)*p, where p is an odd prime and k is in N; then binomial(2*j,j) == 2^j (mod j) if and only if p satisfies the following conditions:
a) p divides binomial(2^(k+1),2^k) - 2^(2^k);
b) p has at least k 1's in its binary expansion.
Theorem. If m is an even perfect number then j = 2m satisfies the congruence binomial(2*j,j) == 2^j (mod j). See A000396.
Theorem. Let j = p^2 with p a prime number. Then p is a Wieferich prime if and only if binomial(2*j,j) == 2^j (mod j). See A001220. (End)
Contains 17179738112 and 274877382656 (from Guedes-Machado paper). - Michael De Vlieger, Nov 22 2023
Contains 3386741824, 750984028672, 33029195197184, 1145067923695616, 422612863956511744. - Ricardo Machado, Nov 23 2023
Contains 84385517065596416, 62648180117928433664, 273984397779878971648, 36506097537257040703232. - Max Alekseyev, Dec 07 2023
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LINKS
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PROG
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(PARI) lista(nn) = {forcomposite(n=1, nn, if (binomod(2*n, n, n) == Mod(2, n)^n, print1(n, ", "))); } \\ Michel Marcus, Dec 06 2013 and Dec 03 2023
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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